What makes Geometric CFT easier than CFT? I've been reading:
math.stanford.edu/~conrad/249BPage/handouts/geomcft.pdf
in an attempt to shed some geometric light on class field theory. The last paragraph there reads:

In case the ground field $k$ is perfect, the essential difficulty in the proof of class field theory – proving
that the Artin map kills certain principal ideals – becomes easy to prove geometrically by means of the
interpretation of geometric points of generalized Jacobians in terms of generalized ideal class groups. (More
precisely, one has $J_m (k) = Cl_m (K)$ when $Br(k) = 1$, as happens when $k$ is finite but not when $k$ is a number
field.)

What precisely is he referring to? What particular part of class field theory ("that the Artin map kills certain principal ideals") becomes easier to prove, and why ("by means of the
interpretation of geometric points of generalized Jacobians in terms of generalized ideal class groups.")?
 A: You may want to read [Milne, Arithmetic Duality Theorems], Chapter I, Appendix A http://jmilne.org/math/Books/ADTnot.pdf and also [Neukirch, Schmidt, Wingberg], Cohomology of Number Fields, Chapter X, §1.
What makes it easier? The existence of the Jacobian and the known structure of its rational points, that everything is a curve over a field, so the theory is more geometric. Also the Zeta function being rational simplifies things.
Edit: Furthermore, the Chebotarev density theorem for function fields is easier to prove than for number fields (see, e.g., Fried, Function Field Arithmetic, Chapter 6.4 [cf. also http://ricerca.mat.uniroma3.it/dottorato/Chebotarev/JardenChebotarevfunctionfields.pdf] vs. 6.5): In the function field case it follows from the Riemann hypothesis for curves.
A: $\def\Spec{\mathrm{Spec}\ }$Let me try to add an answer with a bit more big picture. I'm not an expert here, so I'm hoping Matt E., B. and K. Conrad and others will help improve this. I'll start by outlining the proof of geometric CFT which you'll find in Serre's Algebraic Groups and Class Fields. (There is another line of proof which you'll find in, for example, David Ben Zvi's MSRI talk, and I am still absorbing.)
Theorem 1 Let $L/K$ be a finite Galois extension (with no further conditions on the fields) with $A = \mathrm{Gal}(L/K)$. Then there is an algebraic group $G$ over $K$, an embedding of $A \times \Spec(K)$ into $G$, and maps $\Spec(L) \to G$ and $\Spec K \to G/A$, such that the diagram
$$\begin{matrix}
\Spec L & \rightarrow & G \\ \downarrow & & \downarrow \\ \Spec K & \rightarrow & G/A
\end{matrix}$$
commutes, with $\Spec L$ the preimage of $\Spec K$ within $G$, and with the two actions of $A$ on $\Spec L$ (by multipication in $G$, and by the Galois action) being equal.
This sounds very technical, but you probably have seen two special cases of it. 
Special Case 1 If $K$ contains the $m$-th roots of unity and $A = \mathbb{Z}/m$, then we can take $G$ to be the multipicative group and $A \subset G$ to be the $m$-th roots of unity; the theorem then says that $L$ is of the form $K(\alpha^{1/m})$ for some $\alpha \in K^{\ast}$. This is Kummer's thoerem. The maps $L \to G$ and $K \to G/A \cong G$ are given by $\alpha^{1/m}$ and $\alpha$.
Special Case 2 If $K$ has characteristic $p$ and $A = \mathbb{Z}/p$, then we can take $G$ to be the additive group and $A \subset G$ to be the elements of $\mathbb{F}_p$. Then $G/A \cong G$, with the map $G \to G/A \cong G$ being $y \mapsto y^p-y$. So the theorem says that there is some $\beta \in K$ such that $L = K(y)/(y^p-y-\beta)$. This is the Artin-Schrier theorem. 
If you are familiar with the proofs of Kummer and Artin-Schrier, the proof of Theorem 1 isn't much harder.
Theorem 2 Let $A$, in the above theorem, be abelian. Then we can take $G$ to be abelian as well.
In this case, $G/A$ is itself an algebraic group, which we will call $J$. 
We now specialize to the case that $K$ and $L$ are the fields of meromorphic functions on curves $X$ and $Y$ over some ground field $k$. Write $f: Y \to X$ for the covering map. We do not yet assume that $k$ is finite.
Theorem 3 We can take the group $G$ to be defined over $k$, with $A \subseteq G(k)$.
This means that we can interpret the above diagram more geometrically: We can find an open subset $X^{\circ}$ of $X$ such that, setting $Y^{\circ}:= f^{-1}(X^{\circ})$, there is a commutative diagram
$$\begin{matrix}
Y^{\circ} & \rightarrow & G \\ \downarrow & & \downarrow \\ X^{\circ} & \rightarrow & G/A
\end{matrix}$$
as before. 
Now let $k= \mathbb{F}_q$ and let $F$ denote the $q$-power Frobenius.
Theorem 4 When $A$ is abelian and $k = \mathbb{F}_q$, we can find a map $h: J \to G$ such that the composites $G \to J \to G$ and $J \to G \to J$ are $F-1$. 
Warning I suspect that I am missing a technical hypothesis, perhaps that $L$ and $\bar{k}$ be disjoint.
Special Cases In an Artin-Schrier extension, $G \cong J$ is the additive group $\mathbb{G}_a$ and the map $G \to J$ is $y \mapsto y^p-y$. If $m | q-1$, then $\mathbb{F}_q$ contains the $m$-th roots of unity. In the Kummer extension, $G \cong J \cong \mathbb{G}_m$, with $G \to J$ being $x \mapsto x^m$ and $J \to G$ being $x \mapsto x^{(q-1)/m}$. So the composite is $x \mapsto x^{q} x^{-1}$.
Generalizing the computation in my other answer shows:
Theorem 5 With the above hypotheses, if $D$ is a divisor supported on $X^{\circ}$, and $\bar{D}$ the divisor which it splits into in $X^{\circ}(\bar{k})$, then 
$$h \left( \sum_{\pi \in \bar{D}} \pi \right) = \mathrm{Art}(D)$$.
Here the sum $\sum_{\pi \in \bar{D}} \pi$ lives in $J(\bar{k})$, the map $h$ puts it into $G(\bar{k})$, and the assertion is it is equal to $\mathrm{Art}(D)$ under the embedding $A \to G(k)$.
Examples In the Kummer case, Theorem 5 says that $\mathrm{Art}(D) = \prod_{\pi \in \bar{D}} \alpha(\pi)^{(q-1)/m}$. In the Artin-Schrier case, Theorem 6 says that $\mathrm{Art}(D) = \sum_{\pi \in \bar{D}} \beta(\pi)$.  
Now, for the result which uses the most technical tools. This is basically Theorem 3.2 in Brian Conrad's notes. By the way, this works for any perfect $k$, not just a finite field.
Theorem 6 If $A$ is abelian, we can take $J$ to be a generalized Jacobian $J(\mathfrak{m}, X)$, for some conductor $\mathfrak{m}$, and $X \to J$ we can take a translation of the standard map $X \to J(\mathfrak{m}, X)^1$. Moreover, we can take the support of $\mathfrak{m}$ to be the set $S$ of critical points of $Y \to X$, and we can take $X^{\circ} = X \setminus S$. In particular, if $Y \to X$ is unrammified, we can take $J$ to be the Jacobian and $S = \emptyset$. 
Warning I might be slightly missing something in the last sentence; I would have expected to see conditions like "$Y$ is geometrically connected" and "$Y(k) \neq \emptyset$" showing up.
Let me point out the subtlety of the last sentence of Theorem 6. Let $\mathrm{char}(k) \neq 2$, and let $A = \mathbb{Z}/2$. Then $L = K(\sqrt{\alpha})$ for some $\alpha$. In our high-tech language, we can take $G$ to be the multiplicative group $\mathbb{G}_m$, with $A$ embedded as $\{ \pm 1 \}$. $X^{\circ}$ is the locus where $\alpha$ is nonzero and the map $X \to G/A \cong G$ is given by the function $\alpha$ on $X^{\circ}$. Even if $Y \to X$ is an unbranched cover, the function $\alpha$ will still have zeroes and poles (of even order). We cannot map the entire projective curve $X$ to an affine group like $\mathbb{G}_m$. Theorem 6 is telling us that, by using a very different group, the Jacobian of $X$, we can arrange for the map to $J$ to be defined everywhere on $X$.
Combining Theorems 5 and 6 and the definition of $J(\mathfrak{m}, X)$, we have
Key Theorem Let $k$ be finite and $A$ abelian. Then there is a modulus $\mathfrak{m}$, supported on the ramified primes $S$ of $Y \to X$, such that 
$\mathrm{Art}$ is trivial on principal ideals whose generators are $1 \bmod \mathfrak{m}$.

Now, what happens if we try to work with number fields? There are three problems. To me, Problem 1 feels the most serious.
Problem 1 The best attempt at a generalization of Theorem 3 is to replace $G$ and $J$ with group schemes $\mathcal{G}$ and $\mathcal{J}$ over $\mathcal{O}_K$. So the map $X^{\circ} \to J$ turns into a section of $\mathcal{J} \to \Spec \mathcal{O}_K$ defined on $X^{\circ}$. But this means that different points of $X^{\circ}$ wind up lying in different fibers of the group scheme, so a sum as in Theorem 6 makes no sense.
Problem 2 There is no global map $F$, so it is unclear what an analogue of Theorem 4 would look like.
Problem 3 Although the ray class groups of $K$ generalize the groups $J(\mathfrak{m}, X)(k)$, there is no algebraic group which generalizes $J(\mathfrak{m}, X)$, so it is not clear what an analogue of Theorem 6 would look like.
A: There is a lot here and it is hard to tell where you are confused. I'll try to fill in some steps along the way. For simplicity, I'll present the whole theory only for unramified extensions. I'll assume that you have gone through CFT for number fields in the ideal theoretic (not adelic) presentation, as in Janusz or Cox's books. 
$\def\Art{\mathrm{Art}}$Let $K$ be a number field, and let $L/K$ be an unramified abelian extension with Galois group $A$. Let $I(K)$ be the group of fractional ideals of $K$. So the Artin map is the map $\Art:I(K) \to A$ which sends a prime ideal $\pi$ to the Frobenius lift of $\pi$. 
Key Theorem: With the above notation, $\Art$ is trivial on the principal ideals.
Example: The extension $\mathbb{Q}(\sqrt{-5}, i)/\mathbb{Q}(\sqrt{-5})$ is unramified (exercise!) with Galois group $\{ \pm 1 \}$. Let $\pi$ be a prime of $\mathbb{Q}(\sqrt{-5})$ and, for simplicity, let $\pi \neq \langle 2, \sqrt{-5}-1 \rangle$, which is the unique prime lying over 2. Then $\mathrm{Frob}(\pi)=1$ if $-1$ is square in $\mathbb{Z}[\sqrt{-5}]/\pi$ and $\mathrm{Frob}(\pi)=-1$ if $-1$ is not square in this residue field. So the above theorem says the following: Take any odd element $\alpha \in \mathbb{Z}[\sqrt{-5}]$ and factor the ideal $(\alpha)$ as $\pi_1 \pi_2 \cdots \pi_r$. Then there is an even number of $\pi_k$ for which $-1$ is square in $\mathbb{Z}[\sqrt{-5}]$. In fact, the class group of $\mathbb{Z}[\sqrt{-5}]$ is $\mathbb{Z}/2$, and it turns out that $-1$ is a square in $\mathbb{Z}[\sqrt{-5}]/\pi$ if and only if $\pi$ is principal. 
(I encourage you to check a half dozen examples of this!)
I hope you will agree that this theorem is extremely deep. Even the corollary about $\mathbb{Z}[\sqrt{-5}]$ in the above example is not obvious, although it can be proved by elementary means.
It is the Key Theorem which Brian Conrad is calling the "essential difficulty" in CFT. More precisely,  you need to prove the generalization of this theorem for ramified extensions, which requires the introduction of an auxiliary modulus $\mathfrak{m}$, in which case the set of all principal ideals is replaced by a certain subgroup of principal ideals. But all the key points I want to make can be seen in the unramified setup.
Now, let's see how this Key Theorem, in the geometric case, is deduced from Theorem 3.2 in Conrad's handout. Theorem 3.2 is itself not obvious, but (in my experience) proofs of it are usually more motivated than direct assaults on the Key Theorem.
$\def\bark{\bar{k}}$Here is a specialization of Theorem 3.2 which contains all the cases I want to talk about. Let $k = \mathbb{F}_q$ and $\bark$ be the algebraic closure of $k$. The letter $F$ will always denote the $q$-power Frobenius; what $F$ is acting on will (hopefully) be clear from context. Let $f: Y \to X$ be an unbranched abelian cover of $X$. Assume further that $Y \times_k \bark$ is connected and $Y(k) \neq 0$. These last two conditions aren't really necessary, but they make the exposition simpler. (If you don't want simplifying assumptions, read Conrad!)  
Let $y_0 \in Y(k)$ and let $x_0 \in X(k)$ be the image of $y_0$. Algebraically, $x_0$ is a prime of the downstairs field and $y_0$ is a prime above it with residue extension of degree $1$. Because $Y \to X$ is abelian, all the primes in $g^{-1}(x_0)$ have the same degree; in other words, $g^{-1}(x_0) \subseteq Y(k)$. This will be relevant later.
Let $J$ be the Jacobian of $X$, and map $X \to J$ by $x \mapsto x-x_0$.
Theorem 3.2 specialized: With the above notation, there is a finite unramified abelian cover $g: G \to J$ such that the diagram
$$\begin{matrix} 
Y & \rightarrow & G \\ \downarrow & & \downarrow \\ X & \rightarrow & J 
\end{matrix}$$
commutes, with $Y = X \times_J G$. (Compare this with the diagram in Conrad's Theorem 3.2.)
Sketch of the deduction of the Key Theorem from the above: The following won't be easy, but it is (I hope) easier than standard proofs of CFT.
Claim 1: $G \times_k \bark$ is connected. Proof Sketch: If not, then $Y \times_k \bark$ would also be disconnected.
From the general theory of abelian varieties, this tells us that $G$ is an abelian variety and $J \times_k \bark=(G \times_k \bark)/H$ for a finite subgroup $H$ of $G(\bark)$. 
Claim 2: $H \subseteq G(k)$. Proof sketch We have $H = g^{-1}(0) = f^{-1}(x_0) \subseteq Y(k)$, as discussed above. We have $Y(k) \subseteq G(k)$.
Note that $G(k) = \mathrm{Ker}(F-\mathrm{Id})$, where $F - \mathrm{Id}$ maps $G \to G$. So $H$ is contained in the kernel of $F - \mathrm{Id}$. We deduce that we can factor $G \to J \to G$ where the composite is $F - \mathrm{Id}$. This proof is already getting too long, so let me make things simpler by assuming that the second map $J \to G$ is an isomorphism, and leave adapting the more general case as an exercise for you.
Our setting is now the following: We have a curve $X$ embedded in its Jacobian $J$. $Y$ is the curve
$$\{ y \in J : \mathrm{Frob}(y) - y \in X \}$$
The map $Y \to X$ is $y \mapsto F(y) - y$. Note that this is indeed an abelian cover: The abelian group $J(k)$ acts on $Y$ with the element $z \in J(k)$ acting by $y \mapsto y+z$. So we identify $A = J(k)$. The fact that we have been able to reduce from an arbitrary abelian cover to one of this special form is very deep (and, of course, most of the work is inside Theorem 3.2).
Now, let's understand what the Artin map does. Let $\pi$ be a prime in the downstairs field, with residue field $\mathbb{F}_{q^r}$. Geometrically, $\pi$ corresponds to a set $(\pi_1, \pi_2, \ldots, \pi_r) \in X(\bark)$ such that $F(\pi_i) = \pi_{i+1}$ and $F(\pi_r) = \pi_1$. Let $y \in Y(\bark)$ lie above $\pi_1$. Then the Frobenius lift at $\pi$ is required to send $y$ to $F^r(y)$. So $\mathrm{Frob}_{\pi}$ is the unique element $z$ of $J(k)$ such that $y+z = F^r(y)$. In other words, $z=F^r(y)-y$. 
Now, here comes a key computation:
$$z=F^r(y) - y = \sum_{i=1}^r \left(  F^{i}(y) - F^{i-1}(y) \right) = \sum_{i=1}^r F^{i-1} \left( F(y)-y \right) = \sum_{i=1}^r F^{i-1}(\pi_1) = \sum_{i=1}^r \pi_i$$
So $\mathrm{Frob}(\pi)$ is equal, in $J(\bark)$, to the sum of the points into which $\pi$ splits in $X(\bark)$. 
We deduce the following corollary by linearity: Let $D$ be a divisor in $X$. Let $D$ split in $X(\bark)$ as $\sum a_i \pi_i$, for various points $\pi_i \in X(\bark)$. Then $\Art(D) = \sum a_i \pi_i$, where the sum takes place in $J(\bark)$.
In particular, if $D$ is principal, then $\Art(D)=0$, as desired. QED.
